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Let $P(z)$ be a polynomial of degree $n$ with $|P(z)|\leq 1$ on $|z|=1$ and $P_m(z)$ be a partial sum of $P(z).$ How large $P_m(z)$ can be on $|z|=1?$

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    $\begingroup$ Trivial upper bound is $\|1+e^{i \theta}+...+e^{im \theta}\|_{L^{1}(\mathbb{T})}$. $\endgroup$ May 25 '20 at 4:43
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    $\begingroup$ What here is fixed and what is allowed to vary? $\endgroup$
    – user44191
    May 25 '20 at 4:46
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The trivial upper bound $\max_{|w|=1}|P_{m}(w)|\leq \|1+z+\cdots+z^{m}\|_{L^{1}(\mathbb{T})} \asymp C \log(m)$ that I wrote in the comment is actually sharp in the regime $m=n/2$. Here is the proof.

Notice that $P_{m}(z)$ is convolution of $P(z)$ with $D_{m}(z) = 1+z+\cdots+z^{m}$ on the unit circle, therefore, by the triangle inequality $\max_{|z|=1}|P_{m}(z)| \leq \frac{1}{2\pi}\int_{-\pi}^{\pi}|1+e^{i\theta}+\cdots+e^{i m \theta}| d\theta \asymp C \log(m)$

Next, let us show that the upper bound is sharp in the regime $m=\frac{n}{2}$ where $n$ is large.

Indeed, consider the polynomial $$ P(z) = z^{n} \overline{\left( 1+\frac{z}{1}+\cdots+\frac{z^{n}}{n}\right)} - z^{n}\left(1+\frac{z}{1}+\cdots+\frac{z^{n}}{n} \right) = z^{n}+\frac{z^{n-1}}{1}+\cdots+\frac{1}{n}-z^{n}-\frac{z^{n+1}}{1}-\cdots-\frac{z^{2n}}{n} $$ it is of degree $2n$, and $\max_{|z|=1}|P_{n-1}(z)|\geq |P_{n-1}(1)| \asymp \log(n)$. On the other hand let us show that $\max_{|z|=1}|P(z)|\asymp 1$. Indeed, for $z=e^{ix}$ we have
$$ |P(z)| = 2\left| \sum_{k=1}^{n} \frac{\sin(kx)}{k}\right|\leq 2 \int_{0}^{\pi} \frac{\sin(s)}{s}ds \asymp 1 \quad \text{for all} \quad n \geq 1, \; x \in [0, 2\pi) $$ The last inequality I guess is known "since Democritus".

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    $\begingroup$ @user159888 I do not see typos. One small thing is that instead of P_{n-1} I should be writing P_{n}. But both are fine it just gives example of degree 2n with m=n-1 $\endgroup$ May 26 '20 at 12:43
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    $\begingroup$ I haven't looked closely, but it feels like this argument might symmetrize to get an upper bound of $O(\log(\min(m,n-m)))$? $\endgroup$ May 26 '20 at 15:17
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    $\begingroup$ Yes, this can be done by writing $P_{m}(z) = P(z)-(P(z)-P_{m}(z))$ the first term is estimated by $1$, and the term $(P(z)-P_{m}(z))$ is upper bounded by $\|z^{m+1}+...+z^{n}\|_{L^{1}(\mathbb{T})}\asymp \log(n-m)$. $\endgroup$ May 27 '20 at 2:56

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